# Stress-energy tensor of a perfect fluid

1. Oct 25, 2006

### notknowing

The stress energy tensor of a perfect fluid is composed of two terms of which only one term contains the metric tensor gab. (product of metric tensor and pressure). For curved spacetime, one replaces the flat spacetime metric tensor by the metric tensor of curved space. What I find bizar however is that the metric tensor enters into the expression of the stress-energy tensor in such an asymmetric way (one part affected, the other not). The same applies for the stress energy tensor of the electromagnetic field. Is there some deep reason why one part is affected while the other part is not ? Should one not expect that the geometry of spacetime affects all components in a similar way ?

2. Oct 25, 2006

### pmb_phy

Huh? Where did you get that from? The stress-energy-momentum for a perfect fluid is

T^{\alpha\beta} = \rho U^{\alpha} U^{\beta}

(I used "tex" in square brackets but got a latex error. What did I do wrong??)

where U is the 4-momentum of a fluid element

If you're talking about the components of the tensor at a given point then you can choose whatever system of coordinates you like, even locally a locally flat coordinate system in which the metric tensor is diag(1,-1,-1,-1).

Best wishes

Pete

Last edited: Oct 25, 2006
3. Oct 25, 2006

### MeJennifer

In the wikipedia article http://en.wikipedia.org/wiki/Fluid_solution" [Broken] it is written as:

$$T^{ab} = (\mu + p) \, u^a \, u^b + p \, g^{ab}$$

I think he is asking why the velocity vectors are not subject to the metric as the pressure is.

By the way, happy birthday Pete!

Last edited by a moderator: May 2, 2017
4. Oct 25, 2006

### notknowing

Indeed, my question is why the vectors are not subject to the metric as the pressure is.

Last edited by a moderator: May 2, 2017
5. Oct 25, 2006

### pmb_phy

Thank you Jen! Much appreciated. By the way, how did you know it was my birthday today?

The Birthday Boy Pete :tongue2:

ps - My previous post was incorrect. I posted the tensor for a dust. Thanks Jen for getting it right and correcting me.

Last edited: Oct 25, 2006
6. Oct 25, 2006

### Stingray

You're essentially splitting the stress-energy tensor into two parts that are orthogonal to each other. Since the density portion has the tensor structure $u^{a} u^{b}$, the pressure term should be proportional to a rank-2 symmetric tensor that leaves zero when contracted with the 4-velocity. But contracting it with any vector orthogonal to u will just give you that vector back (with a raised index). It's easy to verify that the only object satisfying these requirements is the standard projection operator
$$g^{ab} + u^{a} u^{b}$$

Edit: TeX doesn't seem to be working right now. The first bit is supposed to read u^{a} u^{b}, and the second should say g^{ab}+u^{a}u^{b}.

Last edited: Oct 25, 2006
7. Oct 26, 2006

### notknowing

I understand that there are clear and strict mathematical rules to construct the tensors with the right properties but is there some way to understand this from a more physical point of view ? How is it possible that part of the tensor (the uaub term) is completely independent of the metric ? Maybe it is not possible to have a better understanding than the mathematical point of view. I was just wondering ..

8. Oct 26, 2006

### Stingray

How could it involve the metric? In a sense, u^{a} involves it implicitly since it is supposed to be normalized. But beyond that, any terms would just be absorbed into \rho.

9. Oct 27, 2006

### notknowing

Sorry for the late reply but more down to earth problems have occupied me
My knowledge of mathematics and relativity is not so advanced as yours, so I have some difficulty in following your arguments. I do not understand how the metric is implicitly involved in u^{a} (by normalisation). Could you explain this in some more detail ?

10. Oct 27, 2006

### Stingray

When writing the $\rho u^{a} u^{b}$, it is assumed that $u^{a} u_{a} = g_{ab} u^{a} u^{b} = -1$. You cannot make this statement without using the metric. If it isn't imposed, $\rho$ loses its meaning.

11. Oct 27, 2006

### notknowing

Thanks for this quick reply. I got your point. Suppose one would multiply the metric by some constant (scalar), how would this change, in general, the stress energy tensor of matter, fields, etc. ?

12. Oct 27, 2006

### George Jones

Staff Emeritus
It's also illustrative to consider Stingray's last point using index-free notation, i.e., consider the stress-energy tensor to be a bilinear map that takes pairs of 4-vectors into scalars. The stress-energy tensor for a fluid acting on arbitrary 4-vectors $v$ and $w$ gives

$$T \left( v , w \right) = \left( \rho + p \right) g \left( v , u \right) g \left( w , u \right) + pg \left( v , w \right).$$

13. Oct 29, 2006

### notknowing

Thanks George. Coming back to my last question ; If one would multiply the metric tensor by some scalar constant A, how would the stress energy tensor of a perfect fluid, or the stress energy tensor in general, be modified ?

14. Oct 29, 2006

### Stingray

Sorry I forgot to answer before. Anyway, what you're asking is a special case of something known as a conformal transformation (or conformal isometry). Multiplying the metric by a positive constant $A^{2}$ is equivalent to a coordinate transformation $x \rightarrow x/A$. The stress-energy tensor would then transform as $T^{ab} \rightarrow T^{ab}/A^{2}$.

15. Oct 29, 2006

### notknowing

Thanks Stingray for this very interesting remark. How did you work out this so quickly ?
What I'm actually trying to find out is how the Einstein Field equations transform when the metric tensor is multiplied by a constant or in other words, how would the Einstein tensor change under such a transformation ?

16. Oct 29, 2006

### Stingray

By the (old fashioned) definition of tensor, a coordinate transformation $\bar{x}^{a}=\bar{x}^{a}(x)$ transforms the components of any tensor with two upper indices as

$$\bar{L}^{ab} = \frac{ \partial \bar{x}^{a} }{ \partial x^{c} } \frac{ \partial \bar{x}^{b} }{ \partial x^{d} } L^{cd}$$

Similarly,

$$\bar{L}_{ab} = \frac{ \partial x^{c} }{ \partial \bar{x}^{a} } \frac{ \partial x^{d} }{ \partial \bar{x}^{b} } L_{cd}$$

This applies to the Einstein and metric tensors just like it does to any other.

If the factor that you're multiplying the metric by isn't a constant, then the interpretation as a coordinate transformation isn't always possible. But it can still be a useful thing to do, and is called a conformal transformation. It is described in detail in one of the appendices in Wald's textbook.

Last edited: Oct 29, 2006
17. Oct 30, 2006

### notknowing

So, if I understand it correctly, the Einstein Field Equations ARE invariant under a multiplication of the metric tensor by a constant. But would this invariance not be spoiled by adding the cosmological constant term ? And would this not be an argument against the introduction of this term ?

18. Oct 30, 2006

### Stingray

Multiplying the metric by a constant is equivalent to a coordinate transformation. Einstein's equations (with or without the cosmological constant) are invariant under all coordinate transformations.

19. Nov 1, 2006

### notknowing

:surprised Something must be wrong here. If you multiply the metric by a constant, then the cosmological constant term will also be multiplied by that same constant, while the stress energy tensor will be divided by that constant (your previous messages). So, clearly they do not scale in the same way and so Einsteins equations can not be invariant under such a change.

20. Nov 1, 2006

### pervect

Staff Emeritus
I don't understand what the problem is. Stingray is right, multiplying by a constant is just a coordinate transformation. Perhaps you need to set up a specific example. Suppose we have a metric g, and a metric gg, representing the same space-time, and all of the coefficients of gg are 4 times as large as the coefficeints of g.

Then ds^2 = g_ij dx^i dx^j = gg_ij dxx^i dxx^j

where gg_ij = 4 g_ij
xx^i = (1/2) x^i

So when you label all coordinates x^i with 1/2 their old labels, you increase the metric coefficeints g_ij by a factor of 4